%!TEX program = xelatex
%!TEX TS-program = xelatex
%!TEX encoding = UTF-8 Unicode

\documentclass[12pt,t,aspectratio=169,mathserif]{beamer}
%Other possible values are: 1610, 149, 54, 43 and 32. By default, it is to 128mm by 96mm(4:3).
%run XeLaTeX to compile.

\input{../wang_slides_preamble_windows.tex}

\begin{document}

\title{Complex Analysis}
\subtitle{Chapter 1. Complex Numbers \\  
Section 2. The Geometric Representation of Complex Numbers}
%\institute{SLUC}
\author{LVA}
%\date
%\renewcommand{\today}{\number\year \,年 \number\month \,月 \number\day \,日}
%\date{ {2023年9月21日} }

\maketitle


%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
% 设置方程编号从16开始
\setcounter{equation}{15} % 因为设置为15，下一个使用的标签就会从16开始

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\begin{frame}{Contents }

\vspace{-0.4cm}\noindent\makebox[\linewidth]{\rule{\paperwidth}{0.4pt}}


\begin{enumerate}

\item  The Algebra of Complex Numbers

\begin{enumerate}
\item[1.1.] Arithmetic Operations
\item[1.2.] Square Roots
\item[1.3.] Justification
\item[1.4.] Conjugation, Absolute Value
\item[1.5.] Inequalities
\end{enumerate}

\item  {\color{red}The Geometric Representation of Complex Numbers}

\begin{enumerate}
\item[2.1.] {\color{red}Geometric Addition and Multiplication}
\item[2.2.] {\color{red}The Binomial Equation}
\item[2.3.] {\color{red}Analytic Geometry}
\item[2.4.] {\color{red}The Spherical Representation}
\end{enumerate}

\end{enumerate}

\end{frame}

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%\begin{frame}{ 作业1A: 7-12 }
%
%\vspace{-0.4cm}\noindent\makebox[\linewidth]{\rule{\paperwidth}{0.4pt}}
%
%\begin{enumerate}
%
%\item[7.] 用平面几何作图画出复数的加法、减法、乘法、除法。
%\item[8.] 求出一个复数关于一条直线对称的另一个复数。
%\item[9.] 求出经过给定三个复数的圆的圆心和半径。
%\item[10.] 证明经过特定两点的圆与单位圆垂直。
%\item[11.] 计算两个复数在球极投影球面上的两个点的距离。
%\item[12.] 证明球极投影球面上的任意直径的两个点对应的两个复数的关系等式。
%
%\end{enumerate}
%
%\end{frame}
%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\begin{frame}{2. The Geometric Representation of Complex Numbers. }

\vspace{-0.4cm}\noindent\makebox[\linewidth]{\rule{\paperwidth}{0.4pt}}

\begin{enumerate}

\item[1.] 
With respect to a given rectangular coordinate system in a plane, the complex number $a = \alpha + i\beta$ can be represented by the point with coordinates $(\alpha,\beta)$. 

\item[2.] 
This representation is constantly used, and we shall often speak of the point $a$ as a synonym of the number $a$.

\item[3.] 
The first coordinate axis ($x$-axis) takes the name of {\color{blue}real axis}, and the second coordinate axis ($y$-axis) is called the {\color{blue}imaginary axis}. 

\item[4.] 
The plane itself is referred to as the {\color{blue}complex plane}. 

\end{enumerate}

\end{frame}

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\begin{frame}{2. The Geometric Representation of Complex Numbers. }

\vspace{-0.4cm}\noindent\makebox[\linewidth]{\rule{\paperwidth}{0.4pt}}

\begin{enumerate}


\item[5.] 
The geometric representation derives its usefulness from the vivid mental pictures associated with a geometric language. 

\item[6.] 
{\color{blue}We take the point of view, however, that all conclusions in analysis should be derived from the properties of real numbers, and not from the axioms of geometry. }

\item[7.] 
For this reason we shall use geometry only for descriptive purposes, and not for valid proof, unless the language is so thinly veiled that the analytic interpretation is self-evident. 

\item[8.] 
This attitude relieves us from the {\color{blue}exigencies} of rigor in connection with geometric considerations.

\end{enumerate}

\end{frame}

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\begin{frame}{2.1. Geometric Addition and Multiplication. }

\vspace{-0.4cm}\noindent\makebox[\linewidth]{\rule{\paperwidth}{0.4pt}}

\begin{enumerate}
\item[1.] 
{\color{red}The addition of complex numbers can be visualized as vector addition.}

\item[2.] 
To this end we let a complex number be represented not only by a point, but also by a vector pointing from the origin to the point. 

\item[3.] 
The number, the point, and the vector will all be denoted by the same letter $a$.

\item[4.] 
As usual we identify all vectors which can be obtained from each other by {\color{blue}parallel displacements}.

\end{enumerate}

\end{frame}

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\begin{frame}{2.1. Geometric Addition and Multiplication. }

\vspace{-0.4cm}\noindent\makebox[\linewidth]{\rule{\paperwidth}{0.4pt}}

\begin{enumerate}

\item[5.] 
Place a second vector $b$ so that its {\color{blue}initial point} coincides with the {\color{blue}end point} of $a$.

\item[6.] 
Then $a + b$ is represented by the vector from the initial point of $a$ to the end point of $b$. 

\item[7.] 
To construct the difference $b - a$ we draw both vectors $a$ and $b$ from the same initial point; then $b - a$ points from the end point of $a$ to the end point of $b$.

\item[8.] 
Observe that $a + b$ and $a - b$ are the diagonals in a {\color{blue}parallelogram} with the sides $a$ and $b$ (Fig. 1-1).

\end{enumerate}

\end{frame}

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\begin{frame}{2.1. Geometric Addition and Multiplication. }

\vspace{-0.4cm}\noindent\makebox[\linewidth]{\rule{\paperwidth}{0.4pt}}

\begin{figure}[ht!]
\centering
\includegraphics[height=0.5\textheight, width=0.75\textwidth]{figure-1-1.png}
%\caption{Fig. 1-1. Conformal Mapping of a polygon }
\end{figure}

\end{frame}

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\begin{frame}{2.1. Geometric Addition and Multiplication. }

\vspace{-0.4cm}\noindent\makebox[\linewidth]{\rule{\paperwidth}{0.4pt}}

\begin{enumerate}

\item[9.] 
An additional advantage of the vector representation is that the length of the vector $a$ is equal to $|a|$. 

\item[10.] 
Hence the distance between the points $a$ and $b$ is $|a - b|$. 

\item[11.] 
With this interpretation the triangle inequality $|a + b| \le |a| + |b|$ and the identity $|a + b|^2 + |a - b|^2 = 2(|a|^2 + |b|^2)$ become familiar geometric theorems. 

\end{enumerate}

\end{frame}

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\begin{frame}{2.1. Geometric Addition and Multiplication. }

\vspace{-0.4cm}\noindent\makebox[\linewidth]{\rule{\paperwidth}{0.4pt}}

\begin{enumerate}

\item[12.] 
The point $a$ and its {\color{blue}conjugate} $\bar{a}$ lie symmetrically with respect to the real axis.

\item[13.] 
The symmetric point of $a$ with respect to the imaginary axis is $-\bar{a}$. 

\item[14.] 
The four points $a$, $-\bar{a}$, $-a$, $\bar{a}$ are the vertices of a rectangle which is symmetric with respect to both axes.


\end{enumerate}

\end{frame}

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\begin{frame}{2.1. Geometric Addition and Multiplication. }

\vspace{-0.4cm}\noindent\makebox[\linewidth]{\rule{\paperwidth}{0.4pt}}

\begin{enumerate}
\item[15.] 
{\color{red}In order to derive a geometric interpretation of the product of two
complex numbers we introduce polar coordinates. }
 
\item[16.] 
If the {\color{blue}polar coordinates} of the point $(\alpha,\beta)$ are $(r,\varphi)$, we know that
\begin{equation*}
\begin{aligned}
\alpha &= r \cos \varphi, \\ 
\beta &= r \sin \varphi. 
\end{aligned}
\end{equation*}

\item[17.] 
Hence we can write $a = \alpha + i\beta = r(\cos \varphi + i \sin \varphi)$. 

\item[18.] 
In this trigonometric form of a complex number $r$ is always $\ge 0$ and equal to the {\color{blue}modulus} $|a|$.

\item[19.] 
The polar angle $\varphi$ is called the {\color{blue}argument} or {\color{blue}amplitude} of the complex number, and we denote it by $\mathrm{arg}\, a$.

\end{enumerate}

\end{frame}

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\begin{frame}{2.1. Geometric Addition and Multiplication. }

\vspace{-0.4cm}\noindent\makebox[\linewidth]{\rule{\paperwidth}{0.4pt}}

\begin{enumerate}

\item[20.] 
Consider two complex numbers $a_1 = r_1(\cos \varphi_1 + i \sin \varphi_1)$ and $a_2 = r_2(\cos\varphi_2 + i \sin \varphi_2)$.

\item[21.] 
Their product can be written in the from 
\begin{equation*}
\begin{aligned}
a_1a_2 = & r_1r_2[(\cos\varphi_1 \cos \varphi_2 - \sin \varphi_1 \sin \varphi_2) \\
& + i(\sin \varphi_1 \cos \varphi_2 + \cos \varphi_1 \sin \varphi_2)]. 
\end{aligned}
\end{equation*}

\item[22.] 
By means of the {\color{blue}addition theorems of the cosine and the sine} this expression can be simplified to
\begin{equation}
a_1a_2 = r_1r_2[ \cos(\varphi_1+\varphi_2) + i\sin(\varphi_1+\varphi_2) ]. 
\label{eq-16}
\end{equation}

\item[23.] 
We recognize that the product has the modulus $r_1r_2$ and the argument $\varphi_1 + \varphi_2$. 

\end{enumerate}

\end{frame}

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\begin{frame}{2.1. Geometric Addition and Multiplication. }

\vspace{-0.4cm}\noindent\makebox[\linewidth]{\rule{\paperwidth}{0.4pt}}

\begin{enumerate}

\item[24.] 
The latter result is new, and we express it through the equation
\begin{equation}
\mathrm{arg}\,(a_1a_2) = \mathrm{arg}\,(a_1) + \mathrm{arg}\,(a_2). 
\label{eq-17}
\end{equation}

\item[25.] 
It is clear that this formula can be extended to arbitrary products, and we can therefore state: 

\item[26.] 
{\color{red}\it The argument of a product is equal to the sum of the arguments of the factors. }

\item[27.] 
This is fundamental.

\item[28.] 
The rule that we have just formulated gives a deep and unexpected justification of the geometric representation of complex numbers. 


\end{enumerate}

\end{frame}

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\begin{frame}{2.1. Geometric Addition and Multiplication. }

\vspace{-0.4cm}\noindent\makebox[\linewidth]{\rule{\paperwidth}{0.4pt}}

\begin{enumerate}

\item[29.] 
We must be fully aware, however, that the manner in which we have arrived at the formula (\ref{eq-17}) violates our principles. 

\item[30.] 
In the first place the equation (\ref{eq-17}) is between angles rather than between numbers, and secondly its proof rested on the use of trigonometry. 

\item[31.] 
{\color{red}Thus it remains to define the argument in analytic terms and to prove (\ref{eq-17}) by purely analytic means.}

\item[32.] 
For the moment we postpone this proof and shall be content to discuss the consequences of (\ref{eq-17}) from a less critical standpoint.


\end{enumerate}

\end{frame}

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\begin{frame}{2.1. Geometric Addition and Multiplication. }

\vspace{-0.4cm}\noindent\makebox[\linewidth]{\rule{\paperwidth}{0.4pt}}

\begin{enumerate}


\item[33.] 
We remark first that the argument of 0 is not defined, and hence (\ref{eq-17}) has a meaning only if $a_1$ and $a_2$ are $\neq 0$.

\item[34.] 
Secondly, the {\color{blue}polar angle} is determined only up to multiples of 360°. 

\item[35.] 
For this reason, if we want to interpret (\ref{eq-17}) numerically, we must agree that multiples of 360° shall not count.


\end{enumerate}

\end{frame}

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\begin{frame}{2.1. Geometric Addition and Multiplication. }

\vspace{-0.4cm}\noindent\makebox[\linewidth]{\rule{\paperwidth}{0.4pt}}

\begin{enumerate}

\item[36.] 
By means of (\ref{eq-17}) a simple geometric construction of the product $a_1a_2$ can be obtained. 

\item[37.] 
{\color{blue} It follows indeed that the triangle with the vertices $0, 1, a_1$ is similar to the triangle whose vertices are $0, a_2, a_1a_2$. } 

\item[38.] 
The points $0, 1, a_1$, and $a_2$ being given, this similarity determines the point $a_1a_2$ (Fig. 1-2).


\end{enumerate}

\end{frame}

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\begin{frame}{2.1. Geometric Addition and Multiplication. \hfill 作业1A-7 }

\vspace{-0.4cm}\noindent\makebox[\linewidth]{\rule{\paperwidth}{0.4pt}}

\begin{figure}[ht!]
\centering
\includegraphics[height=0.6\textheight, width=0.65\textwidth]{figure-1-2.png}
%\caption{Fig. 1-1. Conformal Mapping of a polygon }
\end{figure}

\end{frame}

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\begin{frame}{2.1. Geometric Addition and Multiplication. }

\vspace{-0.4cm}\noindent\makebox[\linewidth]{\rule{\paperwidth}{0.4pt}}

\begin{enumerate}

\item[39.] 
In the case of division (\ref{eq-17}) is replaced by (\ref{eq-18})
\begin{equation}
\mathrm{arg}\,\left(\frac{a_2}{a_1}\right) = \mathrm{arg}\,(a_2) - \mathrm{arg}\, (a_1). 
\label{eq-18}
\end{equation}

\item[40.] 
{\color{red} The geometric construction is the same, except that the similar triangles are now $0, 1, a_1$ and $0, a_2/a_1, a_2$. }

\item[41.] 
Remark: A perfectly acceptable way to define angles and arguments would be to apply the familiar methods of calculus which permit us to express the length of a circular arc as a definite integral.

\item[42.] 
This leads to a correct definition of the trigonometric functions, and to a computational proof of the addition theorems. 


\end{enumerate}

\end{frame}

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\begin{frame}{2.1. Geometric Addition and Multiplication. }

\vspace{-0.4cm}\noindent\makebox[\linewidth]{\rule{\paperwidth}{0.4pt}}

\begin{enumerate}


\item[43.] 
The reason we do not follow this path is that complex analysis, as opposed to real analysis, offers a much more direct approach. 

\item[44.] 
The clue lies in a direct connection between the exponential function and the trigonometric functions, to be derived in Chap. 2, Sec. 5. 

\item[45.] 
Until we reach this point the reader is asked to subdue his quest for complete rigor.


\end{enumerate}

\end{frame}

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\begin{frame}{2.1. Geometric A and M. Exercise - 1 \hfill 作业1A-8}

\vspace{-0.4cm}\noindent\makebox[\linewidth]{\rule{\paperwidth}{0.4pt}}

\begin{itemize}
\item  {\color{red}Question. 
Find the symmetric points of a with respect to the lines which bisect the angles between the coordinate axes.
}

\item  Answer. 
\begin{enumerate}
\item 
 
\item 

\item 

\end{enumerate}

\end{itemize}

\end{frame}

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\begin{frame}{2.1. Geometric Addition and Multiplication. Exercise - 2}

\vspace{-0.4cm}\noindent\makebox[\linewidth]{\rule{\paperwidth}{0.4pt}}

\begin{itemize}
\item  {\color{red}Question. 
Prove that the points $a_1, a_2, a_3$ are vertices of an equilateral triangle
if and only if $a_1^2 + a_2^2 + a_3^2 = a_1a_2 + a_2a_3 + a_3a_1$.
}

\item  Answer. 
\begin{enumerate}
\item 
 
\item 

\item 

\end{enumerate}

\end{itemize}

\end{frame}

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\begin{frame}{2.1. Geometric Addition and Multiplication. Exercise - 3}

\vspace{-0.4cm}\noindent\makebox[\linewidth]{\rule{\paperwidth}{0.4pt}}

\begin{itemize}
\item  {\color{red}Question. 
Suppose that $a$ and $b$ are two vertices of a square. Find the two other vertices in all possible cases. 
}

\item  Answer. 
\begin{enumerate}
\item 
 
\item 

\item 

\end{enumerate}

\end{itemize}

\end{frame}

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\begin{frame}{2.1. Geometric A and M. Exercise - 4 \hfill 作业1A-9 }

\vspace{-0.4cm}\noindent\makebox[\linewidth]{\rule{\paperwidth}{0.4pt}}

\begin{itemize}
\item  {\color{red}Question. 
Find the center and the radius of the circle which circumscribes the triangle with vertices $a_1, a_2, a_3$. Express the result in symmetric form.
}

\item  Answer. 
\begin{enumerate}
\item 
 
\item 

\item 

\end{enumerate}

\end{itemize}

\end{frame}

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\begin{frame}{2.2. The Binomial Equation. }

\vspace{-0.4cm}\noindent\makebox[\linewidth]{\rule{\paperwidth}{0.4pt}}

\begin{enumerate}

\item[1.] 
From the preceding results we derive that the powers of 
$$a = r(\cos \varphi + i \sin \varphi)$$ 
are given by 
\begin{equation}
a^n = r^n(\cos n\varphi + i \sin n\varphi).
\label{eq-19}
\end{equation}

\item[2.] 
This formula is trivially valid for $n = 0$, and since 
$$
a^{-1} = r^{-1}(\cos\varphi - i\sin\varphi) = r^{-1}[\cos(-\varphi) + i\sin(-\varphi)]
$$
it holds also when $n$ is a negative integer. 


\end{enumerate}

\end{frame}

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\begin{frame}{2.2. The Binomial Equation. }

\vspace{-0.4cm}\noindent\makebox[\linewidth]{\rule{\paperwidth}{0.4pt}}

\begin{enumerate}

\item[3.] 
For $r = 1$ we obtain {\color{blue}de Moivre's formula}
\begin{equation}
(\cos \varphi + i \sin \varphi)^n = \cos n\varphi + i \sin n\varphi
\label{eq-20}
\end{equation}
which provides an extremely simple way to express $\cos n\varphi$ and $\sin n\varphi$ in
terms of $\cos \varphi$ and $\sin \varphi$.

\item[4.] 
To find the nth root of a complex number a we have to solve the equation
\begin{equation}
z^n =a.
\label{eq-21}
\end{equation}

\end{enumerate}

\vfill 

%\vspace{-0.4cm}
\noindent\makebox[\linewidth]{\rule{\paperwidth}{0.4pt}}

{\footnotesize 
Abraham de Moivre (born May 26, 1667, Vitry, France -- died Nov. 27, 1754, London) was a French mathematician who was a pioneer in the development of analytic trigonometry and in the theory of probability. 
}

\end{frame}

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\begin{frame}{2.2. The Binomial Equation. }

\vspace{-0.4cm}\noindent\makebox[\linewidth]{\rule{\paperwidth}{0.4pt}}

\begin{enumerate}

\item[5.] 
Supposing that $a \neq 0$ we write $a = r(\cos\varphi + i\sin\varphi)$ and 
$$z = \rho (\cos\theta + i\sin\theta).$$

\item[6.] 
Then (\ref{eq-21}) takes the form 
\begin{equation}
\rho^n(\cos n\theta + i\sin n\theta) = r(\cos \varphi + i\sin\varphi). 
\label{eq-22}
\end{equation}

\item[7.] 
This equation is certainly fulfilled if $\rho^n = r$ and $n\theta = \varphi$. 

\item[8.] 
Hence we obtain the root
$$
z = \sqrt[n]{r}\left( \cos\frac{\varphi}{n} + i \sin\frac{\varphi}{n} \right)
$$
where $\sqrt[n]{r}$ denotes the positive $n$th root of the positive number $r$.

\item[9.] 
But this is not the only solution. 


\end{enumerate}

\end{frame}

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\begin{frame}{2.2. The Binomial Equation. }

\vspace{-0.4cm}\noindent\makebox[\linewidth]{\rule{\paperwidth}{0.4pt}}

\begin{enumerate}

\item[10.] 
In fact, (\ref{eq-22}) is also fulfilled if $n\theta$ differs from $\varphi$; by a multiple of the full angle. 

\item[11.] 
If angles are expressed in radians the full angle is $2\pi$, and we find that (\ref{eq-22}) is satisfied if and only if
\begin{equation*}
\theta = \frac{\varphi}{n} + k\cdot \frac{2\pi}{n},
%\label{eq-}
\end{equation*}
where $k$ is any integer. 

\item[12.] 
However, only the values $k = 0, 1, \cdots, n-1$ give different values of $z$.

\item[13.] 
Hence the complete solution of the equation (\ref{eq-21}) is given by
\begin{equation*}
z = \sqrt[n]{r} \left[ \cos \left( \frac{\varphi}{n} + k\cdot \frac{2\pi}{n} \right) + i \sin\left( \frac{\varphi}{n} + k\cdot \frac{2\pi}{n} \right) \right],
 k = 0, 1, \cdots, n-1. 
%\label{eq-}
\end{equation*}


\end{enumerate}

\end{frame}

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\begin{frame}{2.2. The Binomial Equation. }

\vspace{-0.4cm}\noindent\makebox[\linewidth]{\rule{\paperwidth}{0.4pt}}

\begin{enumerate}

\item[14.] 
There are $n$ nth roots of any complex number $\neq 0$. 

\item[15.] 
They have the same modulus, and their arguments are equally spaced.

\item[16.] 
Geometrically, the $n$th roots are the vertices of a {\color{blue}regular polygon} with $n$ sides. 


\end{enumerate}

\end{frame}

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\begin{frame}{2.2. The Binomial Equation. }

\vspace{-0.4cm}\noindent\makebox[\linewidth]{\rule{\paperwidth}{0.4pt}}

\begin{enumerate}

\item[17.] 
The case $a = 1$ is particularly important. 

\item[18.] 
The roots of the equation $z^n = 1$ are called {\color{blue}$n$th roots of unity}, and if we set
\begin{equation}
\omega = \cos\frac{2\pi}{n} + i\sin \frac{2\pi}{n}
\label{eq-23}
\end{equation}
all the roots can be expressed by $1, \omega, \omega^2, \cdots, \omega^{n-1}$. 

\item[19.] 
It is also quite evident that if $\sqrt[n]{a}$ denotes any $n$th root of $a$, then all the $n$th roots can be expressed in the form $\omega^k\cdot \sqrt[n]{a}, k = 0, 1, \cdots, n-1$.


\end{enumerate}

\end{frame}

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\begin{frame}{2.2. The Binomial Equation. Exercise - 1 \hfill 作业1B-3}

\vspace{-0.4cm}\noindent\makebox[\linewidth]{\rule{\paperwidth}{0.4pt}}

\begin{itemize}
\item  {\color{red}Question. 
Express $\cos 3\varphi$, $\cos 4\varphi$, and $\sin 5\varphi$ in terms of $\cos \varphi$ and $\sin \varphi$.
}

\item  Answer. 
\begin{enumerate}
\item 
 
\item 

\item 

\end{enumerate}

\end{itemize}

\end{frame}

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\begin{frame}{2.2. The Binomial Equation. Exercise - 2}

\vspace{-0.4cm}\noindent\makebox[\linewidth]{\rule{\paperwidth}{0.4pt}}

\begin{itemize}
\item  {\color{red}Question. 
Simplify $1 + \cos\varphi + \cos 2\varphi + \cdots + \cos n\varphi$ and 
$\sin\varphi + \sin 2\varphi + \cdots + \sin n\varphi$.
}

\item  Answer. 
\begin{enumerate}
\item 
 
\item 

\item 

\end{enumerate}

\end{itemize}

\end{frame}

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\begin{frame}{2.2. The Binomial Equation. Exercise - 3}

\vspace{-0.4cm}\noindent\makebox[\linewidth]{\rule{\paperwidth}{0.4pt}}

\begin{itemize}
\item  {\color{red}Question. 
Express the fifth and tenth roots of unity in algebraic form.
}

\item  Answer. 
\begin{enumerate}
\item 
 
\item 

\item 

\end{enumerate}

\end{itemize}

\end{frame}

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\begin{frame}{2.2. The Binomial Equation. Exercise - 4}

\vspace{-0.4cm}\noindent\makebox[\linewidth]{\rule{\paperwidth}{0.4pt}}

\begin{itemize}
\item  {\color{red}Question. 
If $\omega$ is given by 
$$
\omega = \cos\frac{2\pi}{n} + i \sin \frac{2\pi}{n},
$$
prove that
$$
1 + \omega^h + \omega^{2h} + \cdots + \omega^{(n-1)h} = 0
$$
for any integer $h$ which is not a multiple of $n$.
}

\item  Answer. 
\begin{enumerate}
\item 
 
\item 

\item 

\end{enumerate}

\end{itemize}

\end{frame}

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\begin{frame}{2.2. The Binomial Equation. Exercise - 5}

\vspace{-0.4cm}\noindent\makebox[\linewidth]{\rule{\paperwidth}{0.4pt}}

\begin{itemize}
\item  {\color{red}Question. 
What is the value of
$$
1 - \omega^h + \omega^{2h} - \cdots + (-1)^{n-1}\omega^{(n-1)h} ?
$$
}

\item  Answer. 
\begin{enumerate}
\item 
 
\item 

\item 

\end{enumerate}

\end{itemize}

\end{frame}

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\begin{frame}{2.3. Analytic Geometry. }

\vspace{-0.4cm}\noindent\makebox[\linewidth]{\rule{\paperwidth}{0.4pt}}

\begin{enumerate}
\item[1.] 
In classical analytic geometry the equation of a locus is expressed as a relation between $x$ and $y$. 

\item[2.] 
It can just as well be expressed in terms of $z$ and $\bar{z}$, sometimes to distinct advantage. 

\item[3.] 
The thing to remember is that a complex equation is ordinarily equivalent to two real equations; in order to obtain a genuine locus these equations should be essentially the same.

\end{enumerate}

\end{frame}

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\begin{frame}{2.3. Analytic Geometry. }

\vspace{-0.4cm}\noindent\makebox[\linewidth]{\rule{\paperwidth}{0.4pt}}

\begin{enumerate}

\item[4.] 
For instance, the equation of a circle is $|z - a| = r$.

\item[5.] 
In algebraic form it can be rewritten as $(z - a)(\bar{z} - \bar{a}) = r^2$. 

\item[6.] 
The fact that this equation is invariant under complex conjugation is an indication that it represents a single real equation. 

\end{enumerate}

\end{frame}

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\begin{frame}{2.3. Analytic Geometry. }

\vspace{-0.4cm}\noindent\makebox[\linewidth]{\rule{\paperwidth}{0.4pt}}

\begin{enumerate}

\item[7.] 
A straight line in the complex plane can be given by a parametric equation $z = a+ bt$, where $a$ and $b$ are complex numbers and $b \neq 0$; the parameter $t$ runs through all real values. 

\item[8.] 
{\color{red}Two equations $z = a + bt$ and $z = a' + b't$ represent the same line if and only if $a' - a$ and $b'$ are real multiples of $b$. }

\item[9.] 
The lines are parallel whenever $b'$ is a real multiple of $b$, and they are equally directed if $b'$ is a positive multiple of $b$.


\end{enumerate}

\end{frame}

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\begin{frame}{2.3. Analytic Geometry. }

\vspace{-0.4cm}\noindent\makebox[\linewidth]{\rule{\paperwidth}{0.4pt}}

\begin{enumerate}

\item[10.] 
The direction of a directed line can be identified with $\mathrm{arg}\, b$.

\item[11.] 
The angle between $z = a + bt$ and $z = a' + b't$ is $\mathrm{arg}\, b'/b$; observe that it depends on the order in which the lines are named.

\item[12.] 
{\color{red}The lines are orthogonal to each other if $b'/b$ is purely imaginary.}


\end{enumerate}
 
\end{frame}

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\begin{frame}{2.3. Analytic Geometry. }

\vspace{-0.4cm}\noindent\makebox[\linewidth]{\rule{\paperwidth}{0.4pt}}

\begin{enumerate}

\item[13.] 
Problems of finding intersections between lines and circles, parallel or orthogonal lines, tangents, and the like usually become exceedingly simple when expressed in complex form.

\item[14.] 
An inequality $|z - a| < r$  describes the inside of a circle. 

\item[15.] 
Similarly, a directed line $z = a + bt$ determines a {\color{blue}right half plane} consisting of all points $z$ with $\mathrm{Im}\,(z - a)/b < 0$ and a {\color{blue}left half plane} with $\mathrm{Im}\,(z - a)/b > 0$.

\item[16.] 
An easy argument shows that this distinction is independent of the parametric representation.

\end{enumerate}

\end{frame}

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\begin{frame}{2.3. Analytic Geometry. Exercise - 1  \hfill 作业1B-4 }

\vspace{-0.4cm}\noindent\makebox[\linewidth]{\rule{\paperwidth}{0.4pt}}

\begin{itemize}
\item  {\color{red}Question. 
When does $az + b\bar{z} + c = 0$ represent a line?
}

\item  Answer. 
\begin{enumerate}
\item 
 
\item 

\item 

\end{enumerate}

\end{itemize}

\end{frame}

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\begin{frame}{2.3. Analytic Geometry. Exercise - 2 \hfill 作业1B-5 }

\vspace{-0.4cm}\noindent\makebox[\linewidth]{\rule{\paperwidth}{0.4pt}}

\begin{itemize}
\item  {\color{red}Question. 
Write the equation of an ellipse, hyperbola, parabola in complex form.
}

\item  Answer. 
\begin{enumerate}
\item 
 
\item 

\item 

\end{enumerate}

\end{itemize}

\end{frame}

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\begin{frame}{2.3. Analytic Geometry. Exercise - 3}

\vspace{-0.4cm}\noindent\makebox[\linewidth]{\rule{\paperwidth}{0.4pt}}

\begin{itemize}
\item  {\color{red}Question. 
Prove that the diagonals of a parallelogram bisect each other and that the diagonals of a rhombus are orthogonal.
}

\item  Answer. 
\begin{enumerate}
\item 
 
\item 

\item 

\end{enumerate}

\end{itemize}

\end{frame}

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\begin{frame}{2.3. Analytic Geometry. Exercise - 4}

\vspace{-0.4cm}\noindent\makebox[\linewidth]{\rule{\paperwidth}{0.4pt}}

\begin{itemize}
\item  {\color{red}Question. 
Prove analytically that the midpoints of parallel chords to a circle lie on a diameter perpendicular to the chords.
}

\item  Answer. 
\begin{enumerate}
\item 
 
\item 

\item 

\end{enumerate}

\end{itemize}

\end{frame}

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\begin{frame}{2.3. Analytic Geometry. Exercise - 5 \hfill 作业1B-6}

\vspace{-0.4cm}\noindent\makebox[\linewidth]{\rule{\paperwidth}{0.4pt}}

\begin{itemize}
\item  {\color{red}Question. 
Show that all circles that pass through $a$ and $1/\bar{a}$ intersect the circle $|z|=1$ at right angles.
}

\item  Answer. 
\begin{enumerate}
\item 
 
\item 

\item 

\end{enumerate}

\end{itemize}

\end{frame}

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\begin{frame}{2.3. Analytic Geometry. Exercise - 5 }

\vspace{-0.4cm}\noindent\makebox[\linewidth]{\rule{\paperwidth}{0.4pt}}

\begin{figure}[ht!]
\centering
\includegraphics[height=0.7\textheight, width=0.42\textwidth]{figure-1-2-3-Ex-5.png}
%\caption{Fig. 1-1. Conformal Mapping of a polygon }
\end{figure}

\end{frame}


%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\begin{frame}{2.4. The Spherical Representation. }

\vspace{-0.4cm}\noindent\makebox[\linewidth]{\rule{\paperwidth}{0.4pt}}

\begin{enumerate}

\item[1.] 
For many purposes it is useful to extend the system $\mathbb{C}$ of complex numbers by introduction of a symbol $\infty$ to represent infinity. 

\item[2.] 
Its connection with the finite numbers is established by setting $a + \infty = \infty + a = \infty$ for all finite $a$, and $b\cdot \infty = \infty\cdot b = \infty$ for all $b\neq 0$, including $b = \infty$. 

\item[3.] 
It is impossible, however, to define $\infty + \infty$ and $0\cdot\infty$ without violating the laws of arithmetic. 

\item[4.] 
By special convention we shall nevertheless write $a/0 = \infty$ for $a\neq 0$ and $b/\infty = 0$ for $b \neq\infty$.

\end{enumerate}

\end{frame}

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\begin{frame}{2.4. The Spherical Representation. }

\vspace{-0.4cm}\noindent\makebox[\linewidth]{\rule{\paperwidth}{0.4pt}}

\begin{enumerate}

\item[5.] 
In the plane there is no room for a point corresponding to $\infty$, but we can of course introduce an ``ideal'' point which we call the {\color{blue}point at infinity}. 

\item[6.] 
The points in the plane together with the point at infinity form the {\color{blue} extended complex plane}. 

\item[7.] 
We agree that every straight line shall pass through the point at infinity. 

\item[8.] 
By contrast, no half plane shall contain the ideal point.


\end{enumerate}

\end{frame}

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\begin{frame}{2.4. The Spherical Representation. }

\vspace{-0.4cm}\noindent\makebox[\linewidth]{\rule{\paperwidth}{0.4pt}}

\begin{enumerate}


\item[9.] 
It is desirable to introduce a geometric model in which all points of the extended plane have a concrete representative.

\item[10.] 
To this end we consider the unit sphere $S$ whose equation in three-dimensional space is $x_1^2 + x_2^2 + x_3^2 = 1$. 

\item[11.] 
{\color{red} With every point on $S$, except $(0,0,1)$, we can associate a complex number
\begin{equation}
z=\frac{x_1+ix_2}{1-x_3}
\label{eq-24}
\end{equation}
and this correspondence is one to one.
}


\end{enumerate}

\end{frame}

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\begin{frame}{2.4. The Spherical Representation. }

\vspace{-0.4cm}\noindent\makebox[\linewidth]{\rule{\paperwidth}{0.4pt}}

\begin{enumerate}


\item[12.] 
Indeed, from (\ref{eq-24}) we obtain
\begin{equation*}
|z|^2 = \frac{x_1^2+x_2^2}{(1-x_3)^2} = \frac{1+x_3}{1-x_3},
%\label{eq-24}
\end{equation*}
and hence
\begin{equation}
x_3 = \frac{|z|^2-1}{|z|^2+1}. 
\label{eq-25}
\end{equation}

\item[13.] 
Further computation yields
\begin{equation}
\begin{aligned}
x_1 &= \frac{z+\bar{z}}{1+|z|^2} \\  
x_2 &= \frac{z-\bar{z}}{i(1+|z|^2)}. 
\label{eq-26}
\end{aligned}
\end{equation}

\end{enumerate}

\end{frame}

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\begin{frame}{2.4. The Spherical Representation. }

\vspace{-0.4cm}\noindent\makebox[\linewidth]{\rule{\paperwidth}{0.4pt}}

\begin{enumerate}


\item[14.] 
The correspondence can be completed by letting the point at infinity correspond to $(0,0,1)$, and {\color{blue}we can thus regard the sphere as a representation of the extended plane or of the extended number system. }

\item[15.] 
We note that the hemisphere $x_3 < 0$ corresponds to the disk $|z| < 1$ and the hemisphere $x_3 > 0$ to its outside $|z| > 1$. 

\item[16.] 
In function theory the sphere $S$ is referred to as the {\color{blue}Riemann sphere}.

\end{enumerate}

\end{frame}

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\begin{frame}{2.4. The Spherical Representation. }

\vspace{-0.4cm}\noindent\makebox[\linewidth]{\rule{\paperwidth}{0.4pt}}

\begin{enumerate}


\item[17.] 
If the complex plane is identified with the $(x_1,x_2)$-plane with the $x_1$ and $x_2$-axis corresponding to the real and imaginary axis, respectively, the transformation (\ref{eq-24}) takes on a simple geometric meaning. 

\item[18.] 
Writing $z = x + iy$ we can verify that
\begin{equation}
x:y:-1 = x_1:x_2:x_3-1, 
\label{eq-27}
\end{equation}
and this means that the points $(x,y,0)$, $(x_1,x_2,x_3)$, and $(0,0,1)$ are in a 
straight line.

\item[19.] 
{\color{red}Hence the correspondence is a central projection from the center $(0,0,1)$ as shown in Fig. 1-3. It is called a {\color{blue} stereographic projection}.}

\end{enumerate}

\end{frame}

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\begin{frame}{2.4. The Spherical Representation. }

\vspace{-0.4cm}\noindent\makebox[\linewidth]{\rule{\paperwidth}{0.4pt}}


\begin{figure}[ht!]
\centering
\includegraphics[height=0.7\textheight, width=0.75\textwidth]{figure-1-3.png}
%\caption{Fig. 1-1. Conformal Mapping of a polygon }
\end{figure}

\end{frame}

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\begin{frame}{2.4. The Spherical Representation. }

\vspace{-0.4cm}\noindent\makebox[\linewidth]{\rule{\paperwidth}{0.4pt}}

\begin{enumerate}


\item[20.] 
The context will make it clear whether the stereographic projection is regarded as a mapping from $S$ to the extended complex plane, or vice versa.

\item[21.] 
In the spherical representation there is no simple interpretation of addition and multiplication.

\item[22.] 
Its advantage lies in the fact that the point at infinity is no longer distinguished.

\item[23.] 
{\color{blue}It is geometrically evident that the stereographic projection transforms every straight line in the $z$-plane into a circle on $S$ which passes through the pole $(0,0,1)$, and the converse is also true. }

\end{enumerate}

\end{frame}

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\begin{frame}{2.4. The Spherical Representation. }

\vspace{-0.4cm}\noindent\makebox[\linewidth]{\rule{\paperwidth}{0.4pt}}

\begin{enumerate}


\item[24.] 
{\color{red}More generally, any circle on the sphere corresponds to a circle or straight line in the $z$-plane. }

\item[25.] 
To prove this we observe that a circle on the sphere lies in a plane $a_1x_1 + a_2x_2 + a_3x_3 = \alpha_0$, where we can assume that $a_1^2 + a_2^2 + a_3^2 = 1$ 
and $0 \le \alpha_0 < 1$. 

\item[26.] 
In terms of $z$ and $\bar{z}$ this equation takes the form 
$$
\alpha_1(z+\bar{z}) - \alpha_2i(z-\bar{z}) + \alpha_3(|z|^2-1) = \alpha_0(|z|^2+1)
$$
or
$$
(\alpha_0-\alpha_3)(x^2+y^2) - 2\alpha_1x -2\alpha_2y +\alpha_0 + \alpha_3 =0. 
$$

\end{enumerate}

\end{frame}

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\begin{frame}{2.4. The Spherical Representation. }

\vspace{-0.4cm}\noindent\makebox[\linewidth]{\rule{\paperwidth}{0.4pt}}

\begin{enumerate}


\item[27.] 
For $\alpha_0 \neq \alpha_3$ this is the equation of a circle, and for $\alpha_0 = \alpha_3$ it represents a straight line.

\item[28.] 
Conversely, the equation of any circle or straight line can be written in this form.
The correspondence is consequently one to one.


\end{enumerate}

\end{frame}

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\begin{frame}{2.4. The Spherical Representation. }

\vspace{-0.4cm}\noindent\makebox[\linewidth]{\rule{\paperwidth}{0.4pt}}

\begin{enumerate}

\item[29.] 
{\color{red}It is easy to calculate the distance $d(z,z')$ between the stereographic projections of $z$ and $z'$. }

\item[30.] 
If the points on the sphere are denoted by $(x_1,x_2,x_3)$, $(x_1',x_2',x_3')$, we have first
$$
(x_1-x_1')^2 + (x_2-x_2')^2 + (x_3-x_3')^2 = 2 - 2(x_1x_1' + x_2x_2' + x_3x_3').
$$
From (\ref{eq-25}) and (\ref{eq-26}) we obtain after a short computation
$$
x_1x_1' + x_2x_2' + x_3x_3' = \frac{(1+|z|^2)(1+|z'|^2)-2|z-z'|^2}{(1+|z|^2)(1+|z'|^2)}.
$$

\end{enumerate}

\end{frame}

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\begin{frame}{2.4. The Spherical Representation. \hfill 作业1B-7 }

\vspace{-0.4cm}\noindent\makebox[\linewidth]{\rule{\paperwidth}{0.4pt}}

\begin{enumerate}


\item[31.] 
As a result we find that
\begin{equation}
d(z,z') = \frac{2|z-z'|}{\sqrt{(1+|z|^2)(1+|z'|^2)}}. 
\label{eq-28}
\end{equation}

\item[32.] 
{\color{red}
For $z'=\infty$ the corresponding formula is
\begin{equation*}
d(z,\infty) = \frac{2}{\sqrt{1+|z|^2}}. 
%\label{eq-28}
\end{equation*}
}


\end{enumerate}

\end{frame}

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%\begin{frame}{2.4. The Spherical Representation. }
%
%\vspace{-0.4cm}\noindent\makebox[\linewidth]{\rule{\paperwidth}{0.4pt}}
%
%\begin{enumerate}

%Let $z=x+iy$ be a complex number and $(x_1,x_2,x_3)$ be its sphereical representation. 
%
%Then 
%\begin{equation*}
%\left\{
%\begin{aligned}
%x_1 &= \frac{2x}{1+x^2+y^2},\\ 
%x_2 &= \frac{2y}{1+x^2+y^2},\\
%x_3 &= \frac{-1+x^2+y^2}{1+x^2+y^2}. 
%\end{aligned}
%\right.
%\end{equation*}
%
%\end{enumerate}
%
%\end{frame}
%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\begin{frame}{2.4. The Spherical Rep. Exercise - 1 \hfill 作业1B-8 }

\vspace{-0.4cm}\noindent\makebox[\linewidth]{\rule{\paperwidth}{0.4pt}}

\begin{itemize}
\item  {\color{red}Question. 
Show that $z$ and $z'$ correspond to diametrically opposite points on the Riemann sphere if and only if $zz' = -1$.
}

\item  Answer. 
\begin{enumerate}
\item 
 
\item 

\item 

\end{enumerate}

\end{itemize}

\end{frame}

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\begin{frame}{2.4. The Spherical Representation. Exercise - 2}

\vspace{-0.4cm}\noindent\makebox[\linewidth]{\rule{\paperwidth}{0.4pt}}

\begin{itemize}
\item  {\color{red}Question. 
A cube has its vertices on the sphere $S$ and its edges parallel to the coordinate axes. Find the stereographic projections of the vertices.
}

\item  Answer. 
\begin{enumerate}
\item 
 
\item 

\item 

\end{enumerate}

\end{itemize}

\end{frame}

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\begin{frame}{2.4. The Spherical Representation. Exercise - 3}

\vspace{-0.4cm}\noindent\makebox[\linewidth]{\rule{\paperwidth}{0.4pt}}

\begin{itemize}
\item  {\color{red}Question. 
Same problem for a regular tetrahedron in general position.
}

\item  Answer. 
\begin{enumerate}
\item 
 
\item 

\item 

\end{enumerate}

\end{itemize}

\end{frame}

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\begin{frame}{2.4. The Spherical Representation. Exercise - 4}

\vspace{-0.4cm}\noindent\makebox[\linewidth]{\rule{\paperwidth}{0.4pt}}

\begin{itemize}
\item  {\color{red}Question. 
Let $Z, Z'$ denote the stereographic projections of $z, z'$, and let $N$ be the north pole. Show that the triangles $NZZ'$ and $Nzz'$ are similar, and use this to derive 
$$
d(z,z') = \frac{2|z-z'|}{\sqrt{(1+|z|^2)(1+|z'|^2)}}. 
$$
}

\item  Answer. 
\begin{enumerate}
\item 
 
\item 

\item 

\end{enumerate}

\end{itemize}

\end{frame}

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\begin{frame}{2.4. The Spherical Representation. Exercise - 5}

\vspace{-0.4cm}\noindent\makebox[\linewidth]{\rule{\paperwidth}{0.4pt}}

\begin{itemize}
\item  {\color{red}Question. 
Find the radius of the spherical image of the circle in the plane whose center is $a$ and radius $R$.
}

\item  Answer. 
\begin{enumerate}
\item 
 
\item 

\item 

\end{enumerate}

\end{itemize}

\end{frame}

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

\end{document}

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
